question_answer

In a right angled \[\Delta ABC,\,\,\angle A={{90}^{\text{o}}}\] and sides a,b,c are respectively, 5 cm, 4 cm and 3 cm. If a force F has moments 0, 9 and 16 (in N cm) units respectively about vertices A, B and C, the magnitude of F is

A) 3                             

B)        4                             

C) 5                             

D)        9

Correct Answer: C

Solution :

Let\[F=x\hat{i}+y\hat{j}\] As the moment of F about A is 0, F vanquishes through point A. Also, \[3x=9\] and\[4y=16\] \[\Rightarrow \] \[x=3,\text{ }y=4\] \[\therefore \]  \[|F|=\sqrt{{{x}^{2}}+{{y}^{2}}}=5\]